Set of all sequences of real numbers having a finite number of non-zero elements.

Question

In Peter Szekeres's "A Course in Modern Mathematical Physics" on page 64, he introduces a vector space he denotes $\hat{\mathbb{R}}^\infty$:

The set $\hat{\mathbb{R}}^\infty$ of all sequences of real numbers $(a_0, a_1, \ldots)$ having only a finite number of non-zero members $a_i \neq 0$ is a vector space, using the same rules of vector addition and scalar multiplication given for $\mathbb{R}^\infty$ in Example 3.6. The elements of $\hat{\mathbb{R}}^\infty$ are real sequences of the form $(a_0, a_1, \ldots, a_m, 0, 0, 0, \ldots)$.

The last sentence confuses me. If I've interpreted it correctly, it says that any non-zero element must occur within the first finite number of elements of the sequence. Why must all non-zero elements occur within the first finite number of elements of the sequence? I think I understand why the non-zero elements can't occur at the end of the sequence: the sequence is of infinite length and so the "end" isn't well-defined. Is that reasoning sound? But, why can't we have something like $(0, \ldots, a_0, a_1, \ldots, a_m, 0, \ldots)$ where there are infinite 0s before the first non-zero element, followed by infinite 0s after the last non-zero element? If this is valid, then we can think of variations on this theme. For example, $(a_0, 0, \ldots, a_1, \ldots, a_m, 0, \ldots)$ (i.e., a non-zero element, followed by infinite 0 elements, followed by some finite number of non-zero elements, followed by infinite 0 elements) and other such sequences.

To add some final, unfinished thoughts to this question, I expect that the validity of my claims above might rest on whether each of the proposed sequences (e.g., $(0, \ldots, a_0, a_1, \ldots, a_m, 0, \ldots)$) is well-formed, which I believe is tantamount to saying that each sequence is a countable set (each sequence in $\hat{\mathbb{R}}^\infty$ is clearly countable by virtue of being able to be placed in a sequence). The last related piece of information that comes to mind here is that the union of countably-many countable sets is countable. Since $(0, 0, \ldots)$ and $(a_0, a_1, \ldots, a_m)$ are countable perhaps we can use a union to combine them in some way to reach $(0, \ldots, a_0, a_1, \ldots, a_m, 0, \ldots)$. Any elucidation of these thoughts and whether they apply and are valid here would be appreciated.

Answer 1

Szekeres, for now, is using the indicated vector space to parametrize polynomials with real coefficients. Things are likely to clear up if you consider that the sum of two polynomials is another polynomial, multiplying a polynomial by a real number gives another polynomial (with real coefficients). Then his space is just what happens if you don't write the powers of the $x,$ just the coefficients with commas in between.

In this light, the shift operators for these sequences are just: (I) multiply a polynomial by $x,$ which increases the degree and forces a zero constant term; (II) in the other direction, cancel the constant term (if nonzero) and divide the resulting polynomial by $x$

Answer 2

a non-zero element, followed by infinite 0 elements, followed by some finite number of non-zero elements, followed by infinite 0 elements

I prefer "entry" over "element" when order matters, to help me distinguish between sets or similar (which have "elements" and order doesn't matter) and sequences or similar (which have "entries" and order matters).

Since this is a sequence, the indices of these entries would be natural numbers (by definition of "sequence"). Let's say $a_0\ne0$ and then $a_1,\ldots$ are "infinite[ly many] 0 [entries]" but that after all of those is $a_N\ne0$ (and then there are infinitely many 0 entries after that, but that's not really relevant).

This implies that there is a natural number $N$ which is bigger than infinitely many other natural numbers. But since $N$ is a natural number greater than $0$, you could write it with some finite number digits; call that number $D$. That means the largest that $N$ could be is ${10}^D-1$. And so the most natural numbers there could be below $N$ is ${10}^D-2$. Since that's a finite number, this contradicts the idea that $N$ "is bigger than infinitely many other natural numbers".

Therefore, the idea that there are infinitely many entries of the sequence before the entry $a_N$ is impossible.

(As pointed out in the comments, it's possible to have ordered families of numbers that are not sequences which have this sort of property. But to do that would require an ordered index set other than the natural numbers, so it wouldn't count as a "sequence".)

Answer 3

Indexing has nothing to do with this. A sequence, by definition, is a function on the natural numbers. The sequence element $a_i$ means $f(i)$ for some function $f$ that defines the sequence. Since the set $$S=\{i\in\mathbb N: a_i=f(i)\ne 0\}$$ is a finite subset of natural numbers, it has a maximum element $n$. Therefore, if $i\gt 0$, $f(i)=a_i\not\in S$ and so $a_i=0$.